Dhani has 4 cards numbered 1 through 4. He removes 2 cards at random and adds their values. What is the probability that the sum is less than or equal to 5?

7/12
3/5 -------incorrect
3/4 --------incorrect
2/3
It's not 3/5 or 3/4.

The only picks that fit are

1 2, 2 1, 1 3, 3 1, 1 4 , 4 1, 2 3, 3 2

The prob(for each case) = (1/4)(1/3) = 1/12
the prob(your event) = 8/12 = 2/3

To solve this problem, we can consider all the possible combinations of cards that Dhani can remove and find the ones where the sum is less than or equal to 5.

First, let's consider the total number of possible combinations when Dhani removes 2 cards. The total number of combinations would be C(4, 2) which is equal to 6.

Now, let's list down all the possible combinations along with their sums:
1+2 = 3
1+3 = 4
1+4 = 5
2+3 = 5
2+4 = 6
3+4 = 7

Out of these 6 combinations, we can see that only the first 3 combinations have a sum less than or equal to 5.

Therefore, the probability that the sum is less than or equal to 5 is 3 out of 6, which simplifies to 1/2 or 2/4.

So, the correct answer is 2/3.

To solve this problem, we need to consider all the possible combinations of choosing two cards from the four cards numbered 1 through 4, and calculate the sum for each combination.

There are a total of C(4, 2) = 6 ways to choose 2 cards out of 4. Let's list all the possible combinations and their sums:

1 + 2 = 3
1 + 3 = 4
1 + 4 = 5
2 + 3 = 5
2 + 4 = 6
3 + 4 = 7

Out of these 6 combinations, only 3 of them have a sum less than or equal to 5: 1 + 2, 1 + 3, and 1 + 4.

Therefore, the probability of getting a sum less than or equal to 5 is 3/6 or simplifying it, 1/2.

So, the correct answer is 1/2.